Physicists propose measure of macroscopicity; Schrodinger's cat scores a 57

Apr 26, 2013 by Lisa Zyga report
Physicists propose measure of macroscopicity; Schrödinger's cat scores a 57
Expected macroscopicities for various proposed and hypothetical quantum superposition experiments. Credit: arXiv:1205.3447

(Phys.org) —The size of an object can be measured in many ways, such as by its mass, volume, or even the number of atoms it contains. And when it comes to quantum physics, "macroscopic" objects are considered to be larger than "quantum" ones, since the former are usually described by classical laws and the latter by quantum laws. However, physicists have been challenging the boundary between these two realms by performing experiments that show that multiparticle objects can exist in quantum superpositions. But there has been no standard measure of macroscopicity until now, as a team of physicists has proposed that the macroscopicity of an object can be measured in terms of certain parameters of the experiment used to probe its quantum superposition, rather than as a single property of the object itself.

Physicists Stefan Nimmrichter of the University of Vienna, Austria, and Klaus Hornberger of the University of Duisburg-Essen, Germany, have published a paper on the new definition of macroscopicity in a recent issue of Physical Review Letters.

In the past, researchers have often measured an object's macroscopicity in terms of the number of atoms in the object. But different atoms are different sizes since they contain different numbers of , which raises the question of whether macroscopicity should be measured in terms of the total number of an object's protons, neutrons, and electrons, or even some other way entirely. As physicists continue to observe such as superposition in larger objects, a standard measurement of macroscopicity is needed for comparing these results.

Rather than defining macroscopicity solely in terms of an object's composition, Nimmrichter and Hornberger's definition is based on the idea that quantum equations can be modified to make an object's state more classical. If an experiment can rule out some of these modifications, then it describes a larger quantum superposition and a more macroscopic object. The more modifications an experiment rules out, the more macroscopic the object is.

Several factors can help rule out modifications to achieve a large macroscopicity. For example, a superposition with a long coherence time and an object with a large mass both rule out modifications, among other factors. Taken together, all of these parameters yield a single number, μ, on a logarithmic scale that can be used to grade macroscopicity. On this scale, the superposition state of an object has the same macroscopicity as that of a single electron existing in a superposition for 10μ seconds.

Based on this criteria, the physicists gave scores to some recent superposition experiments. The macroscopicity record score so far is 12, which was achieved by Nimmrichter, Hornberger, and their colleagues in 2010 with a molecule containing 356 atoms. The scientists estimate that atomic clusters containing half a million gold atoms could score a 23 on the macroscopic scale, a feat that would be challenging but not out of reach of future technology.

And what would Schrödinger's cat score on the new macroscopicity scale? The physicists calculated that a 4-kg cat, when in a where it sits in two positions spaced 10 cm apart for 1 second, would score a 57. Knowing this number won't make realizing it any easier, though. The physicists explain that this situation is equivalent to an electron existing in a superposition for 1057 seconds, which is about 1039 times the age of the universe.

Explore further: Controlling light on a chip at the single-photon level

More information: Stefan Nimmrichter and Klaus Hornberger. "Macroscopicity of Mechanical Quantum Superposition States." Physical Review Letters. DOI: 10.1103/PhysRevLett.110.160403
Also at arXiv:1205.3447 [quant-ph]

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EyeNStein
3 / 5 (6) Apr 26, 2013
Chance of finding a wad of notes big enough to pay off the national debt outside of China- Scores 58
dan42day
2.6 / 5 (5) Apr 27, 2013
I recently calculated that the U.S. national debt would be enough to pave the entire U.S. Interstate freeway system with $100 bills.
StarGazer2011
1 / 5 (2) Apr 27, 2013
what would quantanized relativity smell like; mathematically speaking? Could relativity be recast as discrete (i.e. planck) values? Would that do the trick?
EyeNStein
1 / 5 (3) Apr 27, 2013
'quantanized relativity' probably smells like a smooth graduation between particle quantization levels: i.e. non existent.
We think of relativity in smoothly continuous equations but particles with discrete properties.
@ StarGazer's question is the important one on understanding the nature of reality though: As discrete particles simultaneously exist as distributed wavefunctions. We don't have a mathematics to encompass both 'co-states' simultaneously yet. So gravity/relativity and QED smell incompatible at present.
johanfprins
1 / 5 (2) Apr 27, 2013
As discrete particles simultaneously exist as distributed wavefunctions. We don't have a mathematics to encompass both 'co-states' simultaneously yet. So gravity/relativity and QED smell incompatible at present.


"Discrete particles" cannot exist simultaneously as distributed waves: If an entity diffracts it MUST be a wave and nothing else but a wave! There are no "particles" .

You are correct to state that gravity/relativity is incompatible with QED. This is so, and will ALWAYS be so since QED is NOT physics but Voodoo. Gravity/relativity blends smoothly with Schroedinger's wave equation, since the latter wave equation is already a relativistic equation.

In fact, even Bohr's quantum-rule for electrons around a nucleus is determined by Einstein's Special Theory of Relativity! When this rule applies for an electron circling a nucleus, the electron must form a stationary Schroedinger wave.

EyeNStein
1 / 5 (3) Apr 27, 2013
@johanfprins
Physics would be a lot easier (and the physical world a lot stranger) if everything were only wavefunctions. And Dirac did indeed bring wavefunctions and relativity together: But that became the root of the 'standard model' of Quantum theory. The fact that these spread out wavefunctions co-manifest as the various bosons and fermion particles in localised space is well demonstrated and quantified at this point in time.
If you have a set of equations to unify these two co-states simultaneously there is a nobel prize waiting for you.
johanfprins
1 / 5 (3) Apr 27, 2013
Physics would be a lot easier if everything were only wavefunctions.


Not wavefunctions but actual electromagnetic waves which can all be derived from Maxwell's equations.

And Dirac did indeed bring wavefunctions and relativity together:


No he did not! He "derived" an incorrect wave equation for a SINGLE electron and when he solved it he found that the electron must end up having an energy of MINUS infinity.

The fact that these spread out wavefunctions co-manifest as the various bosons and fermion particles in localised space is well demonstrated and quantified at this point in time.


No it is NOT!

If you have a set of equations to unify these two co-states simultaneously there is a nobel prize waiting for you.


I have the equations: I only hope that I live long enough to break through the bigotry and get the Nobel Prize! BTW, I should already have received this prize 13 years ago when I PROVED that superconduction can occur up to 400 C.
EyeNStein
2 / 5 (4) Apr 27, 2013
Didn't the electric wave interpretation go out with de Broglie?
The Wavefunction is a root probability distribution function which co-exists with (or collapses to produce) particle like behaviour which locallises their interactions. Which is what we will call particles, at least until we understand Heizenberg better.
You will need to unify the strong force and gravity with the electroweak force and derive the various 'particle forms' to get your Nobel.
johanfprins
1 / 5 (2) Apr 27, 2013
Didn't the electric wave interpretation go out with de Broglie?


How did this "go out" with de Broglie? The de Broglie wavelength is caused by the Lorentz transformation; where the latter results from Maxwell's equations.

The Wavefunction is a root probability distribution function which co-exists with (or collapses to produce) particle like behavior which localizes their interactions.


No it is NOT a probability distribution: It is the electric-field potential of the EM-wave which IS the electron.

Which is what we will call particles, at least until we understand Heizenberg better.


What you call a "particle" is a localized EM-wave: It is always a wave! Heisenberg believed in Voodoo.

You will need to unify the strong force and gravity with the electroweak force and derive the various 'particle forms' to get your Nobel.


I do NOT need to derive "particle forms" since there are no "particles" whatsoever!
johanfprins
1 / 5 (3) Apr 27, 2013
The classification of different forces as the electromagnetic force, the "electro-weak" force and the "strong force" is BS. All these forces are the same electromagnetic force which acts under different boundary conditions that are determined by gravity.

The gravity-field is the "tunneling tails" of the matter waves, which have NOTHING to do with tunneling. Tunneling, as modeled within the mainstream literature, is impossible since it requires negative kinetic-energy. What is called "tunneling" occurs by the borrowing of energy (so called "zero-point" energy, or "vacuum-energy" as advocated by the morons) for an allowed time (delta)E in order to SCALE a potential barrier: NOT to move through it: The latter violates the conservation of energy!
Disproselyte
1 / 5 (1) Apr 28, 2013
Quantum - Classical Transition:
This transition is relative, depending on the state of the considered system and well understood in the framework of scale relativity: e.g.: http://luth2.obsp...irac.pdf
johanfprins
1 / 5 (2) Apr 28, 2013
Quantum - Classical Transition:
This transition is relative, depending on the state of the considered system and well understood in the framework of scale relativity: e.g.: http://luth2.obsp...irac.pdf


I have noted this interesting paper before. It is, however, based on Minkowski's "space-time" which does not exist since x,y,z, and t are not linearly independent coordinates as they must be to define unique space-time distances s between any two space-time points like (x,y,z,t) and (x(p),y(p),z(p),t(p)).

Minkowski's "space-time" violates the rules of mathematics which require that there can be only one point (at the origin) within such a "space-time" for which x=0, y=0, z=0, and t=0. In Minkowski "space-time" there are many such points other than the origin, thus making it impossible to have unique distances s between space-time points.
EyeNStein
1 / 5 (3) Apr 28, 2013
@johanfprins
So presumably you don't believe in special relativity either??
Without Minkowski space-time how do you generate invariant space-time displacements?
Its our naturally perceived universal time which doesn't exist, and doesn't let you define a single invariant origin point for all observers in all frames.
johanfprins
1 / 5 (1) Apr 28, 2013
So presumably you don't believe in special relativity either??


I have no stated that I do not believe in the Lorentz-transformation!

Without Minkowski space-time how do you generate invariant space-time displacements?


You do not; since they do not exist!

Its our naturally perceived universal time which doesn't exist, and doesn't let you define a single invariant origin point for all observers in all frames.


All clocks no matter in which inertial reference frame they are and no matter with which velocities they move relative to one another show EXACTLY the same time at ANY instant in time. When you transform an event from an inertial reference frame into another IRF the transformed difference in time is NOT simultaneously displayed on the clocks but are two different times, where each of these times register simultaneously on all the clocks within gravity-free space. If this is not the case the relative speed of light will not be c within all IRF's.
johanfprins
1 / 5 (2) Apr 28, 2013
Part of the abstract of one of my manuscripts{

The equations of the Lorentz-transformation are derived by assuming that a laser-source emits a unidirectional wave-front from the position in space where and when the origins of two inertial reference-frames IRF=K and IRF=K/, moving with a speed v relative to one another, coincide. It is found that the physics when the light-source is stationary within IRF=K/ only allows a Lorentz-transformation from IRF=K/ to IRF=K: Not the reverse transformation from IRF=K into IRF=K/. Similarly, it is found that the physics when the light-source is stationary within IRF=K, only allows a Lorentz-transformation from IRF=K into IRF=K/: Not the reverse transformation from IRF=K/ into IRF=K. For this reason, contrary to accepted dogma, the clocks within IRF=K and IRF=K/ must keep time at exactly the same rate.
EyeNStein
1 / 5 (4) Apr 28, 2013
You could have just said no to relativity to save typing...
The relative speed of light will always be c within all IRF's. Regardless of the light source IRF or the IRF it is viewed/measured from.
As for MST 'not existing'; as it is a valid mathematical abstraction to create Space-Time displacement invariance: Whether it physically exists is open for debate at 2am in a student bar. Meanwhile MST is a useful concept even if it did give us E=MC2 and indirectly the atomic bomb.
johanfprins
1 / 5 (2) Apr 28, 2013
The relative speed of light will always be c within all IRF's. Regardless of the light source IRF or the IRF it is viewed/measured from.


Where have I stated otherwise?

As for MST 'not existing'; as it is a valid mathematical abstraction to create Space-Time displacement invariance: Whether it physically exists is open for debate at 2am in a student bar.


Epicycles were a "valid mathematical abstraction" but are not physics. MST is not even a "valid mathematical abstraction"!

Meanwhile MST is a useful concept even if it did give us E=MC2 and indirectly the atomic bomb.


Where did MST give us E=mc^2? Mass being energy has NOTHING to do with the validity of MST. MST is not physics nor mathematics at all. It is amazing that Minkowski, who was a professor in mathematics, could postulate a space-time by equating two expressions which are separately and independently equal to zero, and then claim that they are connected by a unique space-time distance! Insanity?
Whydening Gyre
1 / 5 (2) May 01, 2013
I can't wait til someone finds out/proves the Quantum Universe and Classical Universe are just a peak and a trough of the same wave....
EyeNStein
1 / 5 (3) May 02, 2013
@johanfprins
E=mc^2 derives in Minkowski space-time by simply rearranging the expressions for momentum and kinetic energy in that space. That is where the expression for relativistic increased mass of a (fast) moving object comes from too.
These physical effects are demonstrated by experiments. Supporting their derivations.
johanfprins
1 / 5 (2) May 02, 2013
@johanfprins
E=mc^2 derives in Minkowski space-time by simply rearranging the expressions for momentum and kinetic energy in that space. That is where the expression for relativistic increased mass of a (fast) moving object comes from too.
These physical effects are demonstrated by experiments. Supporting their derivations.


E=m*c^2 can be derived simply and directly from Newton's laws by the assumption that moving mass energy is different from stationary mass-energy. Why use Minkowski space which violates the mathematics of linear spaces so that there cannot be invariant four-dimensional space-time distances within Minkowski space?
EyeNStein
1 / 5 (2) May 02, 2013
But isn't that the point? We don't live in "linear spaces" except as a localised tangential Newtonian approximation. MST was devised to provide more generalised, invariant, solutions for time and displacement , and even that is only locally linear if gravity fields are present.
johanfprins
1 / 5 (2) May 02, 2013
But isn't that the point? We don't live in "linear spaces" except as a localised tangential Newtonian approximation.


I disagree! As far as Special relativity is concerned we do live in linear spaces with absolute time, just as Newton has assumed. The only difference is that an event that occurs within a linear space which moves with a speed v relative to you is observed relative to the origin of your linear space at a non-coincident time and a non-coincident position. This can be derived from the Lorentz transformation: See: http://www.cathod...tion.pdf

MST was devised to provide more generalised, invariant, solutions for time and displacement , and even that is only locally linear if gravity fields are present.


The problem is that MST violates the rules of mathematics since it claims that x,y,z,and t are linearly independent coordinates while they are not; as you will see when reading my manuscript quoted above.
EyeNStein
1 / 5 (2) May 02, 2013
If you believe in 'absolute time', then why do particles moving at relativistic velocities live so much longer than they should?
johanfprins
1 / 5 (2) May 03, 2013
If you believe in 'absolute time', then why do particles moving at relativistic velocities live so much longer than they should?


This is modeled and explained in detail in the reference I gave you: i.e. within: http://www.cathod...tion.pdf

It is not possible to post the whole derivation here. Please read the document and then ask pertinent questions about the contents of the document.

In short it is not the clock moving with the muon which keeps slower time, but the time at which the muon is generated within its own inertial reference (in which it is stationary) is different from, and does not coincide with the time as measured relative to earth; even though the clock on earth and the clock with the muon keep exactly the same time. It sounds strange but this is what the Lorentz transformation demands when the speed of light is the same within both reference frames.
EyeNStein
2 / 5 (4) May 03, 2013
"it is not the clock moving with the muon which keeps slower time, but the time at which the muon is generated within its own inertial reference (in which it is stationary) is different.."

This is an utterly implausible suggestion compared to the Special Relativity principle which asserts that time passes at a slower rate for the muon because of its large velocity.
Your assertion that, balancing two similar space-time equations by equating 0=0 is invalid, would only apply if the dimentionality was mismatched (eg. apples=oranges) but that is not the case here.
johanfprins
1 / 5 (2) May 04, 2013
"it is not the clock moving with the muon which keeps slower time, but the time at which the muon is generated within its own inertial reference (in which it is stationary) is different.."

This is an utterly implausible suggestion compared to the Special Relativity principle which asserts that time passes at a slower rate for the muon because of its large velocity.
Your assertion that, balancing two similar space-time equations by equating 0=0 is invalid, would only apply if the dimentionality was mismatched (eg. apples=oranges) but that is not the case here.


It might sound implausible but this is what the Lorentz transformation demands. Thus if the Lorentz transformation is correct this derivation must be correct.

The muon is stationary relative to its own clock within its own inertial reference frame: Thus according to the principle of relativity this clock cannot keep slower time. The lifetime of the muon within its own IRF does not increase as you claim it does!.
ValeriaT
1 / 5 (1) May 04, 2013
@johanprins Why we should read about AWT of Zephir, Lorentz Transform of johanprins or Electric Universe of cantdrive in every thread which has nothing to do with it? The muon time shift has nothing to do with subject and it's off-topic here. In addition, your ideas about it are senile and imbecile and they don't fit the experimental findings. The contemporary colliders must be very carefully designed with respect to the relativistic effects and if their theory would be wrong, they couldn't work at all.
The muon is stationary relative to its own clock within its own inertial reference frame: Thus according to the principle of relativity this clock cannot keep slower time
You apparently never understood, what the special relativity and time dilatation is about.
ValeriaT
1 / 5 (1) May 04, 2013
Here you can have an illustrative picture of it: in this animation the accelerated muon not only runs time slower, but it even runs it slowly in just the way, which provides that the speed of light within of muon's reference frame remains constant (actually invariant) - as special relativity requires. If the time wouldn't slow down for muon in its own reference frame, then this requirement couldn't be maintained anymore. As you can see, the time dilatation for moving objects is therefore the direct consequence of special relativity and you cannot beat the simple logics of this animation with any pile of confused math.
johanfprins
1 / 5 (1) May 04, 2013
@ ValeriaT,

You know less about physics, mathematics and logic than an ape knows about religion; and you are just as stupid. Please leave physics to people who have brains and real self awareness.

If the time wouldn't slow down for muon in its own reference frame, then this requirement couldn't be maintained anymore.
This again illustrates how stupid you are. according to Einstein's first postulate a muon's lifetime must be exactly the same within ANY IRF within which it is stationary. The muon approaching the earth is stationary within its own inertial reference frame so that a clock travelling with it CANNOT keep slower time.

Please stop thinking you can play in the big league: You are just too stupid!

ValeriaT
1 / 5 (1) May 04, 2013
according to Einstein's first postulate a muon's lifetime must be exactly the same within ANY IRF within which it is stationary
And believe it or not - it actually IS. If you would understand the dense aether model, in which the particles are gaining their mass and time dilatation, you would understand it clearly. We discussed it here at least five times already together - should I explain it again for you? You aren't willing to accept even the slightest trace of new information or logics from me - so you'll suffer again, and again.
johanfprins
1 / 5 (2) May 04, 2013
according to Einstein's first postulate a muon's lifetime must be exactly the same within ANY IRF within which it is stationary
And believe it or not - it actually IS.


I know it actually is and must be

If you would understand the dense aether model, in which the particles are gaining their mass and time dilatation, you would understand it clearly.


You do not need your dense aether model to understand it.

We discussed it here at least five times already together - should I explain it again for you?


Please save me another embarrassing display of your stupidity.

You aren't willing to accept even the slightest trace of new information or logics from me - so you'll suffer again, and again.


There is NO NEW information or logic from you! Your AWT is plain bullshit. If you want to prove that it is not so, fit your equations to actual data instead of making silly cartoons which can fit any hallucination of any demented mind!
EyeNStein
1 / 5 (3) May 04, 2013
Hey guys:
This slanging match isn't advancing our understanding of the universe we find ourselves in.
Relativity and Quantum Electrodynamics currently rule as theories of 'how stuff happens'. They fit the data/facts and have passed each test applied in the search for any new (inconsistent) physics such as you guys keep dreaming up. The only inconsistencies I've seen are coming out of our heads: Where facts contradict theories, no matter how beautiful or cherished, no matter how many papers or letters after your names - such theories are wrong. No matter how many ways you find to look sideways with one eye closed at the data.
If your theories are any good they must predict a RESULT which current theory cannot explain.
johanfprins
1 / 5 (2) May 05, 2013
If your theories are any good they must predict a RESULT which current theory cannot explain.


A very naive way to look at physics, In the time of Galileo epicycles explained the motion of planets as seen from the earth equally well as the Copernicum system explained them; but epicycles were wrong physics. In fact the same silly argument you have just now posted were then used against Galileo. The Copernicum model cannot predict a result that the epicycles cannot explain.

It is well known in physics that you can get different models which model the same data. Which one is correct? Friar Ockham suggested that the simplest of these models is usually correct.

The equations on which STR is based (Lorentz transformation) is at present accepted by all to be correct. But the results derived from it can be interpreted in different ways: For example, the "time dilation" formula can be interpreted as being the simultaneous times on two clocks or different non-simultaneous time on both.
johanfprins
1 / 5 (2) May 05, 2013
The real arbiter is of course an experiment which can falsify one of the two interpretations. It is claimed in the literature that by flying clocks around the world proved that two clocks moving linearly relative to one another keep time at different rates

However, the clocks being flown around the world do NOT move linearly relative to the clock on earth. Furthermore the flying clocks are being accelerated and are flying at different heights through the earth's gravity field. The data had to be "corrected" to get the result that the experimenters wanted to get. There is thus no real direct proof that two clocks moving linearly relative to one another do keep time at different rates. In fact, if they do,.this will violate Einstein's postulates on which he based his Special Theory of Relativity: Since each clock is stationary within its own IRF, each clock MUST have the same time rate within its own IRF.

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